Question

If $\cos (A-B)=\frac{3}{5}and\tan A\tan B=2$, then which one of the following is correct?

• A. $\sin (A+B)=\frac{1}{5}$
• B. $\sin (A+B)=\frac{-1}{5}$
• C. $\cos (A–B)=\frac{-1}{5}$
• D. $\cos (A+B)=\frac{-1}{5}$

Answer 1

The correct option is D

$\cos (A+B)=\frac{-1}{5}$

The explanation for the correct option:

Finding the value:

Given,

$\begin{array}{rcl}\cos (A–B)& =& \frac{3}{5}\dots .\left(i\right)\\ \tan A\tan B& =& 2\\ \frac{\sin A}{\cos A}\times \frac{\sin B}{\cos B}& =& \frac{2}{1}\end{array}$

Using componendo and dividendo rule,

$\begin{array}{rcl}& \Rightarrow & \frac{(\sin A\sin B+\cos A\cos B)}{(\sin A\sin B–\cos A\cos B)}=\frac{(2+1)}{(2–1)}\\ \frac{\cos (A–B)}{-\cos (A+B)}& =& 3\\ \frac{-3}{5\cos (A+B)}& =& 3[\because From(i\left)\right]\\ \cos (A+B)& =& \frac{-1}{5}\end{array}$

Hence, the correct option is option (D).